What is the integral of (x-e^x)^2 ?

The integral of (x-e^x)^2 is (x^3)/3-2(e^xx-e^x)+1/2 e^{2x}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x-e^x)^2

\int (x - e^{x})^{2} d x

\frac{x ^{3}}{3} - 2 (e^{x} x - e^{x}) + \frac{1}{2} e^{2 x} + C

Solution steps

\int (x - e^{x})^{2} d x

x - e^{x} = (x + e^{\frac{x}{2}}) (x - e^{\frac{x}{2}})

= \int ((x + e^{\frac{x}{2}}) (x - e^{\frac{x}{2}}))^{2} d x

Simplify

((x + e^{\frac{x}{2}}) (x - e^{\frac{x}{2}}))^{2} : x^{2} - 2 e^{x} x + e^{2 x}

= \int x^{2} - 2 e^{x} x + e^{2 x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int x^{2} d x - \int 2 e^{x} x d x + \int e^{2 x} d x

\int x^{2} d x = \frac{x ^{3}}{3}

\int 2 e^{x} x d x = 2 (e^{x} x - e^{x})

\int e^{2 x} d x = \frac{1}{2} e^{2 x}

= \frac{x ^{3}}{3} - 2 (e^{x} x - e^{x}) + \frac{1}{2} e^{2 x}

Add a constant to the solution

= \frac{x ^{3}}{3} - 2 (e^{x} x - e^{x}) + \frac{1}{2} e^{2 x} + C

y^{^{'}} = 2 (x + y)^{2}

y = \frac{4}{x ^{2}}

\frac{\partial}{\partial x} (x sin (x + y))

y^{^{'}} + e^{x} y = x^{2} y^{2}

y = x^{2}, y = 17 x

What is the integral of (x-e^x)^2 ?
The integral of (x-e^x)^2 is (x^3)/3-2(e^xx-e^x)+1/2 e^{2x}+C